\(\dfrac{63}{64}\)

\(\dfrac{32}{64}+\dfrac{16}{64}+\dfrac{8}{64}+\dfrac{4}{64}+\dfrac{2}{64}+\dfrac{1}{64}=\dfrac{32+16+8+4+2+1}{64}=\dfrac{63}{64}\)

 1/2    + 1/4     + 1/8  + 1/16  + 1/32 + 1/64

 = 32/64 + 16/64 + 8/64 + 4/64 + 2/64 + 1/64

 =                         63/64

\(2A=1+\dfrac{1}{2}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{18}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)

\(A=1-\dfrac{1}{64}\)

\(A=\dfrac{63}{64}\)

giúp em với ạ , em cấn gấp ,  4h em học rồi ạ . Cảm ơn những ai giúp em ạ , em tick luôn ạ

= 63/64 nha

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)

\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)

a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)

\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)

\(A=1-\frac{1}{64}\)

\(A=\frac{63}{64}\)

=32/64+14/64+8/64+4/64+2/64+1/64

=32+14+8+4+2+1/64

=61/64

1/2 +1/4+ 1/8+1/16+1/32+1/64

=1/1×2+1/2×2+1/2×4+1/4×4+1/4+8+1/8×8

=1/1-1/2+1/2-1/2+1/2-1/4+1/4-1/4+1/4-1/8+1/8-1/8

=1-1/8

=7/8

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

\(=\frac{63}{64}\)

\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{8}\)+\(\frac{1}{16}\)+\(\frac{1}{32}\)+\(\frac{1}{64}\)=\(\frac{32}{64}\)+\(\frac{16}{64}\)+\(\frac{8}{64}\)+\(\frac{4}{64}\)+\(\frac{2}{64}\)+\(\frac{1}{64}\)

                                                                        =\(\frac{32+16+8+4+2+1}{64}\)

                                                                        =\(\frac{63}{64}\)

hok tốt

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\left(\frac{1}{2}+\frac{1}{8}\right)+\left(\frac{1}{4}+\frac{1}{16}\right)+\left(\frac{1}{32}+\frac{1}{64}\right)\)

\(=\left(\frac{4}{8}+\frac{1}{8}\right)+\left(\frac{4}{16}+\frac{1}{16}\right)+\left(\frac{2}{64}+\frac{1}{64}\right)=\frac{5}{8}+\frac{5}{16}+\frac{3}{64}\)

\(=\frac{40}{64}+\frac{20}{64}+\frac{3}{64}=\frac{40+20+3}{64}=\frac{63}{64}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

\(=1-\frac{1}{64}\)

\(=\frac{63}{64}\)

Ta có:\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)\(\frac{1}{64}\)

     \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}\)\(+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\)\(\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}\)

     \(=1-\frac{1}{64}\)\(=\frac{63}{64}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)

Cách 1:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1-1/2 + 1/2-1/4 + 1/4-1/8 +1/8-1/16 + 1/16-1/32 + 1/32-1/64

B=1-1/64

B=63/64

Cách 2:

B=1/2+1/4+1/8+1/16+1/32+1/64

B=1/2¹+1/2²+1/2³+1/2⁴+1/2⁵+1/2⁶

2B=1+1/2¹+1/2^2+1/2^3+1/2^4+1/2^5

2B-B=1-1/2^6

B=1-1/64 

B=63/64

 A = 1/2 + 1/4 + 1/8 + ... + 1/1024

2A = 1 + 1/2 + 1/4 + ... + 1/512

2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)

A = 1 - 1/1024

A = 1023/1024

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)

\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)

\(A=\frac{1023}{1024}\)

A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

2 x A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)

2 x A - A = \(\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)

A=\(1-\frac{1}{1024}\)

A=\(\frac{1023}{1024}\)